The following content is copyright (c) 2009-2013 by Goods of the Mind, LLC.

This essential trains for: AMC-10, AMC-12, AIME.

A median line of a triangle is a segment that connects a vertex to the midpoint of the opposite side.

Since any two non-parallel lines in a plane will intersect, two of the medians of a triangle will have a common point. Let us denote it with G:

What information do we have about G?

Since the points N and M are midpoints, the segment NM is parallel to the side AC.

Since the segment NM connects midpoints, the triangle NBM is similar to ABC in a ratio of 1:2.

This means that the length of NM is half that of AC.

Another segment can be constructed so as to have the same length as NM by connecting the midpoints of the segments AG and CG.

Since the segments KL and MN are parallel and congruent, the figure KLMN is a parallelogram. Hence, its diagonals bisect each other and we have:

In conclusion, the points K and G divide the median AM in three congruent parts. Similarly, the points L and G divide the median CN in three congruent parts.

Therefore, the point G divides each median in a ratio of 1:2.

Using a transitivity argument, we can show that, by using another pair of medians, they would also intersect at G since the point that divides any median in a given ratio is unique.

Therefore, the medians of a triangle are concurrent and they intersect each other in ratios of 1:2.

The point of intersection of the medians is called centroid or center of gravity of the triangle.

The area property of the medians is based on the fact that, if two triangles share a vertex and the sides opposite from it are collinear, the areas of the triangles are in the same ratio as the lengths of the collinear sides.

In the figure, the blue triangle and the yellow triangle have the same area, which is half of the area of the triangle ABC:

Considering the center of gravity G, the triangles AGC and AGN have areas that are in a ratio of 1:2:

Therefore, the ratio between the area of the triangle AGN and that of ABC is 1:6:

This argument can be repeated to prove that the six areas into which the three medians divide a triangle are equal.

This shows that the area of the triangle is uniformly distributed around the center of gravity. If the triangle is made of a material with constant thickness and density then the point of intersection of the medians is the actual center of mass/gravity of the object.